Question

Heat required in converting $1g$ of ice at $-{10}^{\circ }C$ into steam at ${100}^{\circ }C$ is:

latent heat of fusion= 80 cal/g80 cal/g
latent heat of vaporization=540 cal/g

• A. $3045J$
• B. $6056J$
• C. $721J$
• D. $6J$

Answer 1

The correct option is A $3045J$

We know specific heat of ice is $0.5cal/gK$, specific heat of water=1cal/gK
Also latent heat of fusion= $80cal/g$
latent heat of vaporization=$540cal/g$
The following processes occur during the above said conversion.
$Ice\left(263K\right)\to ice\left(273K\right)\to water\left(273K\right)\to water\left(373K\right)\to steam\left(373K\right)$
Thus total heat absorbed=$(0.5\times 10+80+1\times 100+540)cal=725cal=3045J$